If kinetic energy increases does pressure decrease?

1 answer:

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If molecules are in a closed container then we expect the pressure to increase as the kinetic energy increases. This is because the atoms of an element collide with the walls of the container and increase the pressure.

If we use the formula $PV=nRT$ , where P is the pressure, V is the volume, n is the number of moles, R the ideal gas constant and T is the temperature. According to the formula, P is directly proportional to temperature. An increase in temperature leads to an increase in pressure.

Since we know that temperature is the average kinetic energy of molecules present. It means as we increase the temperature we increase the kinetic energy of the molecules which in turn leads to an increase in the pressure.

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Balancing the reaction by oxidation number method k2cr2o7+sncl2+hcl

Mumz [18]

Answer:

$K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)$

Explanation:

The products of this reaction are given by:

$K_2Cr_2O_7 (aq) + SnCl_2 (aq) + HCl (aq)\rightarrow KCl (aq) + SnCl_4 (aq) + CrCl_3 (aq) + H_2O (l)$

Firstly, dichromate anion becomes chromium(III) cation, let's write this change:

$Cr_2O_7^{2-} (aq)\rightarrow Cr^{3+} (aq)$

The following steps should be taken:

balance the main element, chromium: multiply the right side by 2 to get 2 chromium species on both side:

$Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq)$

balance oxygen atoms by adding 7 water molecules on the right:

$Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

balance the hydrogen atoms by adding 14 protons on the left:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

balance the charge (the total net charge on the left is 12+, on the right we have 6+, so 6 electrons are needed on the left):

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

Similarly, tin(II) cation becomes tin(IV) cation:

$Sn^{2+} (aq)\rightarrow Sn^{4+} (aq) + 2e^-$

Now that we have the two half-equations, multiply the second one by 3, so that it also has 6 electrons that will be cancelled out upon addition of the two half-equations:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

$3 Sn^{2+} (aq)\rightarrow 3 Sn^{4+} (aq) + 6e^-$